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All celestial bodies have a gravitational well, and particle in the vicinity of this well would feel a gravitational force. My questions are:

a)How can I find the thickness of such a gravitational well?

b)Shouldn't the subatomic particles, e.g. An electron be confined to the potential well of a planet or a star?

c)Is quantum mechanical tunnelling a possibility for the electron (here) to get over the potential barrier?

E.g.: A black hole of infinite mass in the presence of another body becomes completely transparent quantum mechanically ( $\Pi$ = 1). But in an Aharonov-Bohm-like effect, if we consider two systems, each with black hole (B.H.) and concentric shell, opposite each other can result in a tunneling probability, $\Pi$ greater than 0.

In a simplified model of a black hole facing a body of mass $M_{2}$. $M_{2}$ is centered at $R$ opposite a black hole of mass $M$ centered at the origin. Since tunneling is greatest near the top of the barrier, the deviation from a $\frac{1}{r}$ potential toward the center of each body is not critical. The potentials used are that of two point masses, so $M_{2}$ may also be a black hole. Thus two little black holes may get quite close for maximum tunneling radiation. Solving the Schrödinger equation outside the black hole:

$$-\frac{\hbar^{2}}{2m}{D^{2}\psi}=-\left[\frac{-GmM}{r}+\left(-\frac{GmM_{2}}{R-r}\right)-E\right]\psi$$ In the region $a\leq r\leq b$, where $a$ bad $b$ are classical turning points, and $$E=-\frac{GmM}{a}+\left(\frac{-GmM}{R-a}\right)=-\frac{GmM}{b}+\left(\frac{-GmM_{2}}{R-b}\right)$$ Since, $T=e^{-2\Delta \gamma}$, $$\Delta \gamma=\frac{m}{\hbar}\sqrt{\frac{2GM}{d}}\left[\sqrt{b(b-d)}-\sqrt{a(a-d)}-dln\left|\frac{\sqrt{b}+\sqrt{b-d}}{\sqrt{a}+\sqrt{a-d}}\right|\right]$$ Here $d=\frac{Ma(R-a)R}{[M((R-a)R+M_{2}a^{2}]}$ and for $R>>b$ & $M_{2}>>M$

Thus, $\Delta \gamma$ approaches zero as $a$ approaches $b$, yielding $\Pi$ approaches 1. When $M$ approaches zero, or $M_{2}$ approaches infinity, or equivalently $\frac{M}{M_{2}}$ approaches zero, $\Delta \gamma$ approaches zero and $\Pi$ approaches one. Hence observing quantum tunnelling.

For a better detailed derivation refer here

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    $\begingroup$ Naveen, I mean no offence to you in any way, but changing the question after an answer has been posted is not really the proper way to ask it. $\endgroup$ – user108787 Nov 21 '16 at 21:16
  • $\begingroup$ @CountTo10 I have first asked you what I added to the question(as an edit) in the comment section under your answer. I hope I didn't offend you and disrespect your answer in any way, my apologies. I am going to edit my question again, but am not adding any new subquestion, just going to mathematically represent what I have highlighted in my question; just wanted to let you know before I do it. $\endgroup$ – Naveen Balaji Nov 21 '16 at 21:22
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    $\begingroup$ No offence taken, I want us both to learn more about the topic, which is the most important thing. But other people might be annoyed if they don't see the full question first ( I honestly don't mind about rep points or things like that), if they feel you don't ask the full question in one go. If you look at the meta posts you might see what I mean. Anyway, all the best :) $\endgroup$ – user108787 Nov 21 '16 at 21:30
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Quantum tunneling enables the penetration of a particle through a potential barrier without the expenditure of energy. It requires that you have the same energy available on both sides of the barrier. In the gravitational "potential well" there is no potential barrier that can be penetrated. Even at infinity the potential is higher than any energy state in the interior of the well.

A gravitational potential barrier exists between two close celestial bodies, like between the moon and the earth. However, these barriers are so extremely thick that a tunneling probability for a subatomic particle can be excluded.

However, a subatomic particle, like an electron, can, due to its low mass, acquire such a high thermal velocity, the escape velocity $$v=\sqrt{\frac{2GM}{r}}$$ that it can (classically) leave the potential well of a celestial body. For the earth at sea level the escape velocity is $11.2km/s$. The mean thermal velocity of an electron at room temperature is $75km/s$. Thus an electron with mean thermal velocity, if uninhibited, would classically leave the potential well of the earth.

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  • $\begingroup$ But what if there are two black holes next to each other, this can result in a tunnelling probability greater than 0. Thanks! $\endgroup$ – Naveen Balaji Nov 21 '16 at 21:11
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    $\begingroup$ +1 but I think we ended up answering different questions :). Anyway, Wikipedia has this line about energy In quantum mechanics, these particles can, with a very small probability, tunnel to the other side, thus crossing the barrier. Here, the "ball" could, in a sense, borrow energy from its surroundings to tunnel through the wall or "roll over the hill", paying it back by making the reflected electrons more energetic than they otherwise would have been but in your answer you say without expenditure of energy. Should it really be, without using extra energy? Regards $\endgroup$ – user108787 Nov 21 '16 at 21:20
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    $\begingroup$ @CountTo10 - Explanations of tunneling using the uncertainty between energy and time are only heuristic. When you solve the time-independent Schrödinger equation for a simple square barrier tunneling problem, you a priori assume wave solutions corresponding to a fixed energy below the barrier height and obtain reflection and transmission probabilities for the waves. The wave solutions in the barrier are purely damped exponentials. Regards.. $\endgroup$ – freecharly Nov 21 '16 at 22:04
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    $\begingroup$ @Naveen Balaji - If the classical gravitation potentials are assumed to be correct, then you can use these potentials in the Schroedinger equation to calculate the tunneling. $\endgroup$ – freecharly Nov 22 '16 at 6:04
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    $\begingroup$ @Naveen Balaji - Sorry, I don't know any paper on QM tunneling in connection with GR. The problem is that in GR gravity is not a force with a potential. But perhaps in special situations it can be equivalently viewed that way. The Schrödinger equation can always be used when you have a potential energy for your particle. $\endgroup$ – freecharly Nov 22 '16 at 6:20
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I was asking myself the same question. Upon further thinking I had concluded that yes it should be possible for an object to tunnel 'out' of such a gravity potential.

As others have pointed out, the gravitational potential well is not a super well defined well. The potential energy increases non-linearly to distance. But this should not mean that tunnelling it is not possible.

One could just simplify the problem into a 2-dimensional gravitational force vs distance problem. In this case the gravitational attractive force must be defined as a function of distance. A Hamiltonian can then be constructed to represent the potential energy of the objects as a function of distance. A particle-like localised wavefunction representing the probability of the particle being found can then be appropriately defined. Then, to see what happens to such a particle over time, one can plot the wavefunction over a range of time (present to future) by letting it evolve over time according to the Hamiltonian.

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How can I find the thickness of such a gravitational well?

Theoretically you shouldn't, the effect of the mass should propagate for ever. Practically, it depends on how sensitive your measuring instruments are.

Shouldn't the subatomic particles, e.g. An electron be confined to the potential well of a planet or a star?

Why? You refer yourself in your title to quantum tunnelling, which is usually used in connection with electromagnetic forces, but it's not exclusively confined to that force. Although, the heavier the mass, the less likely you are to find a particle relatively far away from it. But I think you might be mixing up electrostatic and gravitational wells when you use the word tunnelling in the context of this question.

If there is a well defined thickness, then is tunnelling a possibility for the electron (here) to get over the potential barrier?

There is no abrupt cut off point in these situations. No well defined thickness is involved. Again, I would not see it in conventional tunnelling terms.

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  • $\begingroup$ So would the particles inside the gravitational well not feel the influence of the potential? Also could you please explain a little more on why there must be no thickness associated to the gravitational well. $\endgroup$ – Naveen Balaji Nov 21 '16 at 20:32
  • $\begingroup$ Sure, take the Sun. We can detect the influence of it all the way to the edges of the solar system, it's just that it get weaker, but because a gravity source potentially is infinite in distance, ( since gravitons are massless), it should be felt for ever. Look up the LIGO experiments on Wikipedia to see how weak, but also how far, gravitational effects can be felt. A thickness implies is a wall, with a well defined boundary, but forces such as electromagnetism, ( we can see galaxies a long, long way from us) and gravity do not act as if they have thickness, like a material object has. $\endgroup$ – user108787 Nov 21 '16 at 20:42
  • $\begingroup$ A black hole of infinite mass in the presence of another body becomes completely transparent quantum mechanically ( Π = 1). But in an Aharonov-Bohm-like effect, if we consider two systems, each with black hole (B.H.) and concentric shell, opposite each other can result in a tunneling probability, Π greater than 0. Thanks! $\endgroup$ – Naveen Balaji Nov 21 '16 at 21:00
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    $\begingroup$ @Naveen Balaji - A black hole doesn't have an "infinite mass". Also, this is not necessary. The problem is more that you have to consider two close black holes in the framework of General Relativity which doesn't attribute a potential or force to gravitation. Whether there is something similar in GR to a potential barrier that can be tunneled through is prima facie not clear. $\endgroup$ – freecharly Nov 21 '16 at 21:48

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